Question: $ \int_2^5 \int_0^{4 - \scriptsize\dfrac{4}{3}(x - 2)} dy \, dx$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_2^4 \int_0^{5 - \scriptsize\dfrac{3}{4}y} dx \, dy$ (Choice B) B $ \int_0^4 \int_2^{5 - \scriptsize\dfrac{3}{4}y} dx \, dy$ (Choice C) C $ \int_2^5 \int_0^{4 - \scriptsize\dfrac{3}{4}y} dx \, dy$ (Choice D) D $ \int_0^4 \int_0^{4 - \scriptsize\dfrac{3}{4}y} dx \, dy$
The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $2 < x < 5$ and $0 < y < 4 - \dfrac{4}{3}(x - 2)$. Therefore: ${2}$ ${4}$ ${2}$ ${4}$ ${\llap{-}2}$ $y$ $x$ Because we're switching bounds to $dx \, dy$, we need to start with numeric bounds for $y$. We see that $0 < y < 4$. Then we can define $x$ in terms of $y$. Thus, $2 < x < 5 - \dfrac{3}{4}y$. In conclusion, the double integral after switching bounds is: $ \int_0^4 \int_2^{5 - \scriptsize\dfrac{3}{4}y} dx \, dy$